147 (1995)
A remark concerning random walks with random potentials
by
Yakov G. S i n a i (Princeton, N.J., and Moscow)
Abstract. We consider random walks where each path is equipped with a random weight which is stationary and independent in space and time. We show that under some assumptions the arising probability distributions are in a sense uniformly absolutely con- tinuous with respect to the usual probability distribution for symmetric random walks.
We consider random walks on the d-dimensional lattice Zd with each path having a random statistical weight. Paths starting at (x, k) and ending at (y, n) will be denoted by ωx,ky,n, i.e. ωx,ky,n = {ω(t) ∈ Zd, k ≤ t ≤ n , ω(k) = x, ω(n) = y, kω(t + 1) − ω(t)k = 1}. To define a random weight introduce a sequence of iid rv F = {F (x, t)}, x ∈ Zd, t ∈ Z. Without any loss of generality we may assume that the F (x, t) are given for all x ∈ Zd, t ∈ Z.
The space of all possible realizations of F is denoted by Φ. The measure corresponding to F is denoted by Q, the expectation with respect to Q is denoted by M . We do not use any special notation for the natural σ-algebra in Φ. Our main assumption concerning the distribution of F (x, t) is
M exp(2F (x, t)) < ∞.
The natural group of space-time translations acting in Φ is denoted by {Tx,t}. It preserves the measure Q.
We shall consider the statistical weight of ωx,ky,n equal to π(ωx,ky,n) = exp
nXn
t=k
F (t, ω(t))
o 1
(2d)n−k. Introduce partition functions
Zx,ky,n= X
ωy,nx,k
π(ωy,nx,k), Zx,kn =X
y
Zx,ky,n.
1991 Mathematics Subject Classification: Primary 34F05.
[173]
Now we may define the “random” probability distribution PF ; x,kn defined on paths ωy,nx,k by the formula
p(ωy,nx,k) = π(ωy,nx,k) Zx,kn . The induced probability distribution of y = ω(n) is
py,nx,k =X
ωy,nx,k
p(ωx,ky,n) = Zx,ky,n Zx,kn . We shall also need the usual transition probabilities
q(n−k)(y − x) = X
ωy,nx,k
1 (2d)n−k.
It is well known that for any A > 0 and y for which ky − xk ≤ A√ n, q(n−k)(y − x)
= 1
(2π(n − k)/d)d/2exp
−dky − xk2 2(n − k)
(1 + γ(n−k)(y − x)), where k · k is the Euclidean norm and γ(n−k)(z) tends to zero uniformly in z satisfying the above-mentioned restrictions.
Our purpose in this note is to study the behavior of the distribution of the normalized displacement
ω(n) − ω(k)
p(n − k)/d = y − x p(n − k)/d
with respect to PF ; x,kn as n → ∞. The problem was considered by J. Imbrie and T. Spencer [3] and later by E. Bolthausen [1]. In [1] and [3], it was shown that if the F (x, t) are small enough in appropriate sense, and d ≥ 3, then the limiting distribution of the displacement is Gaussian and for typical F the mean of the square of displacement grows proportionally to time. Recently these results were extended to some random processes with continuous time by J. Conlon and P. Olsen [2]. All these results can be formulated also in terms of diffusion of directed polymers in random environments.
We show below that some of the results of [1] and [3] remain valid under weaker assumptions on the distribution of F and the dimension d. Define
αd=X
n>0
X
z
(q(n)(z))2. This is finite if d ≥ 3. Put
Λ = M exp{F (x, t)} and λ = M exp{2F (x, t)} − Λ2
Λ2 .
Our main assumption is
(1) λαd< 1.
It is easy to see that (1) is valid for d ≥ 3 if λ is small enough. If F (x, t) takes two values ±c with probability 1/2, then (1) is valid for those d for which αd< 1, and does not require the smallness of c. Indeed, in this case always λ < 1, i.e.,
M exp{2F (x, t)} ≤ 2Λ2, because this is equivalent to the obvious inequality
1
2(e2c+ e−2c) ≤ 2
ec+ e−c 2
2 .
If the F (x, t) have Gaussian distribution N (0, σ), then (1) is valid for small enough σ.
Put
h(x, t) = exp{F (x, t)} − Λ Λ
and introduce the series ϕ(x, k) = X
r≥1
X
k≤k1<...<kr
X
z1,...,zr
q(k1−k)(z1− x)q(k2−k1)(z2− z1) . . . . . . q(kr−kr−1)(zr− zr−1)h(z1, k1)h(z2, k2) . . . h(zr, kr), ψ(y, n) = X
r≥1
X
k1<...<kr≤n
X
z1,...,zr
q(k2−k1)(z2− z1) . . .
. . . q(kr−kr−1)(zr− zr−1)q(n−kr)(y − zr)h(z1, k1) . . . h(zr, kr).
It is clear that ϕ(x, t) and ψ(y, t) constitute stationary (with respect to space-time translations) random fields, i.e. ϕ(x, k) = Tx,kϕ(0, 0) and ψ(y, n)
= Ty,nψ(0, 0). Also they are transformed into each other by reversal of time in random walks. This implies, in particular, that the distributions of ϕ(x, t) and ψ(x, t) coincide.
Below we prove the following theorems.
Theorem 1. If (1) is valid then the series giving ϕ(x, k) and ψ(y, n) converge in the space L2(Φ, Q).
Theorem 2. If (1) is valid and ky − xk ≤ A√
n − k where A is any constant, then the partition function Zx,ky,n has the representation
Zx,ky,n = Λn−k+1q(n−k)(y − x)[(1 + ϕ(x, k))(1 + ψ(y, n)) + δ(y,n)(x,k)], where M δx,ky,n = 0 and M (δy,nx,k)2 → 0 as n → ∞, x, k remain fixed and y satisfies the above-mentioned restriction.
P r o o f o f T h e o r e m 1. It is clear that ϕ and ψ are represented as sums of orthogonal vectors in the space L2(Φ, Q). Therefore
M ϕ2(x, k) = X
r≥1
λr X
k<k1<...<kr
X
z1,...,zr
(q(k1−k)(z1− x))2
× (q(k2−k1)(z2− z1))2. . . (q(kr−kr−1)(zr− zr−1))2
= X
r≥1
(λαd)r < ∞.
The same is true for ψ(x, t). We also have M ϕ(x, k) = M ψ(y, n) = 0.
Theorem 2 is proven in Appendix 1.
Theorem 3. If (1) holds then 1 + ϕ(x, t) > 0 and 1 + ψ(y, t) > 0 for Q-a.e. F .
P r o o f. We already showed that M ϕ(x, t) = M ψ(x, t) = 0, M ϕ2(x, t)
> 0 and M ψ2(x, t) > 0. It is enough to consider ϕ(x, k) since ϕ(x, k) and ψ(y, n) have the same distribution. By Theorem 2,
Zx,ky,n
Λn−k+1 − q(n−k)(y − x)[(1 + ϕ(x, k))(1 + ψ(y, n))] = δ(y,n)(x,k)q(n−k)(y − x).
Take a continuous non-negative function f with compact support on Rd, and write
X
y
Zx,ky,n Λn−k+1f
x − y
√n − k
√d
= (1 + ϕ(x, k))X
y
q(n−k)(y − x)f
y − x
√n − k
√d
(1 + ψ(y, n))
+X
y
q(n−k)(y − x)f
y − x
√n − k
√d
δ(y,n)(x,k).
Theorem 2 immediately implies that the last term tends to zero in L2(Φ, Q) for any fixed x, k and n → ∞. Since M ψ(y, n) = 0 the sum
X
y
q(n−k)(y − x)f
y − x
√n − k
√d
(1 + ψ(y, n))
converges in L2(Φ, Q) to C =R
e−kzk2/2f (z) dz/(2π)d/2 > 0. Thus
l.i.m.
n→∞
1 C
X
y
Zx,ky,n Λn−k+1f
y − x
√n − k
√d
= 1 + ϕ(x, k).
Now we can use the obvious inequality Zx,k−2y,n ≥ X
hx,x0i
1 2d
2
eF (x,k−2)+F (x0,k−1)Zx,ky,n= g(x, k − 2)Zx,ky,n, where the last expression gives also the definition of g(x, k − 2) which is positive a.e., and the sum is taken over x0 such that kx − x0k = 1. We use the notation hx, x0i for the nearest neighbors on the lattice. Thus we have (2) 1 + ϕ(x, k − 2) ≥ g(x, k − 2)(1 + ϕ(x, k)).
Assume that 1 + ϕ(x, k − 2) = 0 with positive probability. Take x and consider the set H+ of those numbers 2k such that 1 + ϕ(x, 2k) > 0. It follows from (2) that if 2k ∈ H+ then 2k − 2 ∈ H+. Therefore H+= 2Z1for a.e. F . The ergodicity of T0,2 implies that Q({F : 1 + ϕ(x, k) = 0}) = 0.
Let the conditions of Theorem 2 be valid. As in the proof of Theorem 3 take a continuous function f on Rd with compact support. Using Theorem 2 we can write
(3) X
y
f
y − x
√n − k
√d
Zx,ky,n Zx,kn
= (1 + ϕ(x, k))Λn−k+1 Zx,kn
× X
y
f
y − x
√n − k
√d
q(n−k)(y − x)(1 + ψ(y, n))
+X
y
f
y − x
√n − k
√d
δ(y,n)(x,k)q(n−k)(y − x)
.
Our estimations during the proof of Theorem 2 in the Appendix give l.i.m.
n→∞
Zx,kn
Λn−k+1 = 1 + ϕ(x, k).
Also the last sum in (3) tends to zero in L2(Φ, Q) as n → ∞. Therefore, we have the following theorem.
Theorem 4.
l.i.m.
n→∞
1 Zx,kn
X
y
f
y − x
√n − k
√d
Zx,ky,n=R
f (z)e−kzk2/2 dz (2π)d/2.
This theorem shows in what sense the normalized displacement (ω(n) − ω(k))√
d/√
n − k has the limiting Gaussian distribution. Its variance is the same as for the usual random walk.
Appendix
P r o o f o f T h e o r e m 2. We have Zx,ky,n = X
ωy,nx,k
exp nXn
t=k
F (t, ω(t))
o 1
(2d)n−k
= X
ωy,nx,k
Yn t=k
(Λ + exp{F (t, ω(t))} − Λ) 1 (2d)n−k
= Λn−k+1X
ωx,ky,n
Yn t=k
(1 + h(ω(t), t)) 1 (2d)n−k
= Λn−k+1 h
q(n−k)(y − x) +X
r≥1
X
k≤k1<...<kr<n
X
z1,...,zr
q(k1−k)(z1− x)
× q(k2−k1)(z2− z1) . . . q(kr−kr−1)(zr− zr−1)
× q(n−kr)(y − zr)h(z1, k1) . . . h(zr, kr)h(y, n) i
. In what follows we only deal with the finite sum Zex,ky,n= X
r≥1
X
k≤k1<...<kr≤n
X
z1,...,zr
q(k1−k)(z1− x)q(k2−k1)(z2− z1) . . . . . . q(kr−kr−1)(zr− zr−1)q(n−kr)(y − zr)h(z1, k1) . . . h(zr, kr).
It is clear that M eZx,ky,n= 0 and M ( eZx,ky,n)2= X
r≥1
λr X
k≤k1<...<kr≤n
X
z1,...,zr
(q(k1−k)(z1− x))2
× (q(k2−k1)(z2− z1))2. . . (q(n−kr)(y − zr))2. Fix some constant B whose value will be chosen later and consider
Zex,ky,n(1) = X
r≤B ln n
X
k≤k1<...<kr≤n
X
z1,...,zr
q(k1)(z1− x)
× q(k2−k1)(z2− z1) . . . q(kr−kr−1)(zr− zr−1)
× q(n−kr)(y − zr)h(z1, k1) . . . h(zr, kr).
Let eZx,ky,n(2) be a similar sum where r > B ln n. Then the trivial estimation gives
M ( eZx,ky,n(2))2≤ X
r>B ln n
(λαd)r = (λαd)B ln n 1 − λαd .
Take B so large that (λαd)B ln n
1 − λαd ≤ 1
n2d for all large enough n.
We can write Zx,ky,n
Λn−k+1 = q(n−k)(y − x)(1 + eZx,ky,n(1) + eZx,ky,n(2)).
From our estimations it follows that (i) for all y with ky − xk ≤ A√
n − k the ratio eZx,ky,n(2)/q(n−k)(y − x) tends to zero in L2(Φ, Q) uniformly in y;
(ii) for any continuous function f with compact support, the sum X
y
f
y − x p(n − k)/d
Zey,nx,k(2)
converges to zero in L2(Φ, Q).
Thus it remains to study eZx,ky,n(1) assuming ky − xk ≤ A√
n − k. Let us call an interval (kj−1, kj) large if kj−kj−1≥ nβ for some β with 1/2 < β < 1.
Here k0 = k, kr+1 = n. If r ≤ B ln n then at least one large interval in the sequence (0, k1, k2, . . . , kr, n) is present. We shall show that the main contribution to eZx,ky,n(1) comes from r-tuples (k1, k2, . . . , kr) with only one large interval. Write
Zex,ky,n(1, 1)
= X
0≤r1≤B ln n 0≤r2≤B ln n 1≤r=r1+r2≤B ln n
X
k≤k1<...<kr≤n (kr1,kr1+1) is the unique
large interval
X
z1,...,zr
q(k1)(z1− x)
× q(k2−k1)(z2− z1) . . . q(kr1+1−kr1)(zr1+1− zr1)
× q(kr1+2−kr1+1)(zr1+2− zr1+1) . . . q(kr−kr−1)(zr− zr−1)q(n−kr)(y − zr)
× [h(z1, k1) . . . h(zr1, kr1)] · [h(zr1+1, kr1+1) . . . h(zr, kr)].
We can write Zex,ky,n(1, 1)
q(n)(y − x) = (1 + ϕ(x, k))(1 + ψ(y, n)) − 1 + δ(x,k)(y,n)(2).
The last formula also implies the definition of δ(x,k)(y,n)(2). Since we can re- strict ourselves by summation over those (z1, . . . , zr) where kzr1− xk ≤ n2β, kzr1+1−yk ≤ n2β, the summation over all other z is exceedingly small. Thus M (δx,ky,n(2))2→ 0 as n → ∞ uniformly over all y under consideration.
The rest of our argument is to show that the contribution of r-tuples where the number of large intervals is greater than 1 is relatively small.
Again we write down the square of the norm of the corresponding sum:
Sx,ky,n= αd
X
r≥1
(αdλ)r X
k≤k1<...<kr≤n
X
z1,...,zr
p(k1−k)(z1− x)
× p(k2−k1)(z2− z1) . . . p(kr−kr−1)(z2− z1) . . . . . . p(kr−kr−1)(zr− zr−1)p(n−kr)(y − zr),
where p(i)(z) = (q(i)(z))2/αd. The last double sum can again be considered as the probability that the sum ~η1+. . .+~ηrtakes the values y−x, n−k, where
~ηj = (zj − zj−1, kj − kj−1). It is easy to show that the distribution of the time component of ηj decays as const/td/2. Direct probabilistic arguments show that the probability to have at least two values of j for which the value of the “time” component is greater than nβ decays as 1/n(β+1)d. This shows that the contribution of terms with two large increments (kj − kj−1) to Sx,ky,n(1) is small in L2(Φ, Q) compared with the norm of q(n−k)(y − x).
We omit the details.
I thank K. M. Khanin and Yu. I. Kifer for useful discussions.
The financial support from NSF (grant DMS-9404437) and from the Rus- sian Foundation of Fundamental Research (grant N93-01-16090) are highly appreciated.
References
[1] E. B o l t h a u s e n, A note on the diffusion of directed polymers in a random environ- ment, Comm. Math. Phys. 123 (1989), 529–534.
[2] J. G. C o n l o n and P. A. O l s o n, A Brownian motion version of directed polymer problem, preprint, University of Michigan, 1994.
[3] J. Z. I m b r i e and T. S p e n c e r, Diffusion of directed polymers in a random environ- ment, J. Statist. Phys. 52 (1988), 609–626.
MATHEMATICS DEPARTMENT LANDAU INSTITUTE OF THEORETICAL PHYSICS
PRINCETON UNIVERSITY MOSCOW, RUSSIA
PRINCETON, NEW JERSEY 08540 U.S.A.
Received 5 December 1994;
in revised form 30 January 1995
